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(6x-10)2x=2
We move all terms to the left:
(6x-10)2x-(2)=0
We multiply parentheses
12x^2-20x-2=0
a = 12; b = -20; c = -2;
Δ = b2-4ac
Δ = -202-4·12·(-2)
Δ = 496
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{31}}{2*12}=\frac{20-4\sqrt{31}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{31}}{2*12}=\frac{20+4\sqrt{31}}{24} $
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